Let for i = 1, 2, 3,p_i(x) be a polynomial of | vedclass

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(D) Let $p_i(x) = a_i x^2 + b_i x + c_i$ for $i = 1, 2, 3$. Then $p'_i(x) = 2a_i x + b_i$ and $p''_i(x) = 2a_i$.The matrix $A(x)$ is given by:$A(x) =

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does not depend on $x$

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(D) Let $p_i(x) = a_i x^2 + b_i x + c_i$ for $i = 1, 2, 3$. Then $p'_i(x) = 2a_i x + b_i$ and $p''_i(x) = 2a_i$.The matrix $A(x)$ is given by:$A(x) = \begin{bmatrix} a_1 x^2 + b_1 x + c_1 & 2a_1 x + b_1 & 2a_1 \ a_2 x^2 + b_2 x + c_2 & 2a_2 x + b_2 & 2a_2 \ a_3 x^2 + b_3 x + c_3 & 2a_3 x + b_3 & 2a_3 \end{bmatrix}$.We know that $\det(B(x)) = \det([A(x)]^T A(x)) = \det([A(x)]^T) \det(A(x)) = (\det(A(x)))^2$.Observe the columns of $A(x)$. Let $C_1, C_2, C_3$ be the columns of $A(x)$.$C_1 = \begin{bmatrix} a_1 x^2 + b_1 x + c_1 \ a_2 x^2 + b_2 x + c_2 \ a_3 x^2 + b_3 x + c_3 \end{bmatrix} = x^2 \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix} + x \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix} + \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix}$.$C_2 = \begin{bmatrix} 2a_1 x + b_1 \ 2a_2 x + b_2 \ 2a_3 x + b_3 \end{bmatrix} = 2x \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix} + \begin{bmatrix} b_1 \ b_2 \ b_3 \end{bmatrix}$.$C_3 = \begin{bmatrix} 2a_1 \ 2a_2 \ 2a_3 \end{bmatrix} = 2 \begin{bmatrix} a_1 \ a_2 \ a_3 \end{bmatrix}$.Since $C_3$ is a constant vector,and $C_2$ is a linear combination of $C_3$ and a constant vector,and $C_1$ is a linear combination of $C_3$,$C_2$,and a constant vector,the columns are linearly dependent on $x$ in a way that the determinant $\det(A(x))$ is a constant. Specifically,performing column operations $C_1 	o C_1 - \frac{x}{2} C_2 + \frac{x^2}{4} C_3$ shows that the determinant is independent of $x$.Thus,$\det(A(x))$ is a constant,and consequently,$\det(B(x)) = (\det(A(x)))^2$ is also a constant. Therefore,it does not depend on $x$.

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